Let = { ; for all integers z} be defined as the "orbit" of x. Where means f(f(f(f...(f(x)) n times.

It is not hard to see then that both E and are made of disjoint orbits.

If E or its compliment contains countable many orbits, then the entire set is countable and thus has measure 0.

Suppose m(E) > 0 and E contains uncountably many orbits.

Let S = {one point from each unique orbit}, then S is not measurable. So is not measurable. But , so does it follow that m( ) = 0?

Apr 18th 2010, 01:28 PM

Opalg

This result is essentially the fact that irrational rotations are ergodic. The best way to prove it is by using Fourier series.

Let f be the characteristic function of E,
Extend the domain of f from [0,1] to by making it periodic with period 1. The Fourier coefficients of f are given by . The fact that E is invariant under a translation through a tells you that for all n. But a is irrational, so can never be equal to 1 unless n=0. Therefore for all , which tells you that f is (almost everywhere equal to) a constant function. Since the constant can only be 0 or 1. Thus m(E) must be 0 or 1.

Apr 18th 2010, 05:57 PM

southprkfan1

Quote:

Originally Posted by Opalg

This result is essentially the fact that irrational rotations are ergodic. The best way to prove it is by using Fourier series.

Let f be the characteristic function of E,
Extend the domain of f from [0,1] to by making it periodic with period 1. The Fourier coefficients of f are given by . The fact that E is invariant under a translation through a tells you that for all n. But a is irrational, so can never be equal to 1 unless n=0. Therefore for all , which tells you that f is (almost everywhere equal to) a constant function. Since the constant can only be 0 or 1. Thus m(E) must be 0 or 1.

Unfortunately, I am not familiar with the Fourier series or the term "ergodic," so I cant really relate to this at all. Thanks for the time and effort though.

Quote:

Let S = {one point from each unique orbit}, then S is not measurable. So is not measurable. But , so does it follow that ?

Yes, it does follow, and I think I now know why:

So, I proved that is made up of a bunch of unique orbits.

Now, look at

By translation invariance, we have:

, which is 0 if it is a 0 set and if not. The latter is obviously inadmissible.